Calculate $\lim_{x\to0}\frac{\sin ax\cos bx}{\sin cx}$ 
Calculate: $$\lim_{x\to0} \frac{\sin ax\cos bx}{\sin cx}$$

My attempt:
$$=\lim_{x\to 0} \dfrac {\sin (ax). \cos (bx)}{\sin (cx)}$$
$$=\lim_{x\to 0} \dfrac {\sin (ax)}{ax} \times \dfrac {ax}{\sin (cx)} \times cx \times \dfrac {\cos (bx)}{cx}$$
$$=1\times ax \times 1 \times \dfrac {\cos (bx)}{cx}$$
How do I do further?
 A: A bit of better than your:
$$\lim_{x\rightarrow0}\frac {\sin ax\cos bx}{\sin cx}=\lim_{x\rightarrow0}\left(\cos{bx}\cdot\frac{\sin{ax}}{ax}\cdot\frac{cx}{\sin{cx}}\cdot\frac{a}{c}\right)=\frac{a}{c}$$
A: There is a $0×\infty$ indeterminate form in $1×ax×1×\frac{\cos bx}{cx}$, which would make further resolution a little harder. A more direct approach:
$$\lim_{x\to0}\frac {\sin ax\cos bx}{\sin cx}=\lim_{x\to0}\frac {\sin ax}{\sin cx}$$
$$=\frac{ax-\mathcal O(x^3)}{cx-\mathcal O(x^3)}=\frac ac$$
A: There's nothing wrong with the other answer, but I like to use Taylor series for these kinds of problems, since it is easy to take the limit of a rational function. Here's a few more words about how that expansion works.
Using Taylor series, we can write the following:
$$\lim_{x \to 0} \frac{\sin (ax) \cos(bx)}{\sin(cx)} = \lim_{x \to 0}\frac{(ax - E_{\sin}(x))(1-b^2x^2+E_{\cos}(x))}{cx  - E_{\sin}(x)}$$
Where the $E$ are error functions which go to $0$ as $x$ goes to $0$, and in particular, go to $0$ faster than $x$ does. Technically the top and bottom $E_{\sin}$ are different, but since they are going to $0$ in a moment anyway, we won't clutter things with any more notation.
Multiplying the limit out:
$$\lim_{x\to 0}\frac{ax - ab^2x^3 -E_{\sin}(x)(1-b^2x^2 +E_{\cos}(x))}{cx - E_{\sin}(x)}$$
Now since the error terms go to $0$ faster than $x$, we can cancel an $x$ from every term, and $\frac{E}{x} \to 0$ as $x \to 0$.
$$\lim_{x \to 0} \frac{a - ab^2x^2}{c}$$
Now the limit is obviously $a/c$, as found in the other answer.
A: $$\lim_{x \rightarrow 0}\frac{\sin \text ax \cos \text bx}{\sin \text cx}$$
Using linear approximation
$$L=f(\text a)+f'(x)(x-\text a)$$
Therefore, at $x=0$
$$\sin \text ax\approx \text ax$$
$$\sin \text cx\approx \text cx$$
$$\cos bx \approx 1$$
$$\lim_{x \rightarrow 0}\frac{\sin \text ax \cos \text bx}{\sin \text cx}=\lim_{x \rightarrow 0}\frac{(\text ax)(1)}{\text cx} \iff \lim_{x \rightarrow 0}\frac{(\text ax)(1)}{\text cx}=\lim_{x \rightarrow 0} \frac{\text a}{\text c}$$
$$\therefore \lim_{x \rightarrow 0} \frac{\text a}{\text c}=\frac{\text a}{\text c}$$
